package com.togo.algorithm.medium.design;

import java.util.LinkedList;

/**
 * @Author taiyn
 * @Description 请定义一个队列并实现函数 max_value 得到队列里的最大值，要求函数max_value、push_back 和 pop_front 的均摊时间复杂度都是O(1)。
 * <p>
 * 若队列为空，pop_front 和 max_value 需要返回 -1
 * <p>
 * 示例 1：
 * <p>
 * 输入:
 * ["MaxQueue","push_back","push_back","max_value","pop_front","max_value"]
 * [[],[1],[2],[],[],[]]
 * 输出: [null,null,null,2,1,2]
 * 示例 2：
 * <p>
 * 输入:
 * ["MaxQueue","pop_front","max_value"]
 * [[],[],[]]
 * 输出: [null,-1,-1]
 *  
 * <p>
 * 限制：
 * <p>
 * 1 <= push_back,pop_front,max_value的总操作数 <= 10000
 * 1 <= value <= 10^5
 * @Date 9:17 AM 2023/5/25
 **/
public class Offer59_2 {

    LinkedList<Integer> l1;
    LinkedList<Integer> l2;

    public Offer59_2() {
        l1 = new LinkedList<>();
        l2 = new LinkedList<>();
    }

    public int max_value() {
        if (l2.isEmpty()) return -1;
        return l2.getFirst();
    }

    public void push_back(int value) {

        while(!l2.isEmpty() && l2.getLast() < value)
            l2.removeLast();

        l2.addLast(value);
        l1.addLast(value);
    }

    public int pop_front() {
        if (l1.isEmpty()) return -1;
        int i = l1.removeFirst();
        if (i == l2.getFirst()) {
            l2.removeFirst();
        }

        return i;
    }

    public static void main(String[] args) {
        Offer59_2 obj = new Offer59_2();
        obj.push_back(1);
        obj.push_back(2);
        System.out.println(obj.max_value());
        obj.pop_front();
        System.out.println(obj.max_value());
    }
}
